a. Solve the quadratic equation by factoring. u 2 − 2 u − 35 = 0 b. Solve the equation by using substitution. ( w 2 − 6 w ) 2 − 2 ( w 2 − 6 w ) − 35 = 0
a. Solve the quadratic equation by factoring. u 2 − 2 u − 35 = 0 b. Solve the equation by using substitution. ( w 2 − 6 w ) 2 − 2 ( w 2 − 6 w ) − 35 = 0
Solution Summary: The author calculates the solution for the equation, u2-2u-35=0 by factoring.
a. Solve the quadratic equation by factoring.
u
2
−
2
u
−
35
=
0
b. Solve the equation by using substitution.
(
w
2
−
6
w
)
2
−
2
(
w
2
−
6
w
)
−
35
=
0
Formula Formula A polynomial with degree 2 is called a quadratic polynomial. A quadratic equation can be simplified to the standard form: ax² + bx + c = 0 Where, a ≠ 0. A, b, c are coefficients. c is also called "constant". 'x' is the unknown quantity
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Q.1) Classify the following statements as a true or false statements:
a. If M is a module, then every proper submodule of M is contained in a maximal
submodule of M.
b. The sum of a finite family of small submodules of a module M is small in M.
c. Zz is directly indecomposable.
d. An epimorphism a: M→ N is called solit iff Ker(a) is a direct summand in M.
e. The Z-module has two composition series.
Z
6Z
f. Zz does not have a composition series.
g. Any finitely generated module is a free module.
h. If O→A MW→ 0 is short exact sequence then f is epimorphism.
i. If f is a homomorphism then f-1 is also a homomorphism.
Maximal C≤A if and only if is simple.
Sup
Q.4) Give an example and explain your claim in each case:
Monomorphism not split.
b) A finite free module.
c) Semisimple module.
d) A small submodule A of a module N and a homomorphism op: MN, but
(A) is not small in M.
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